Questions: Hydraulic Machinery: Pumps and Turbines

Question 1

A pump is rated at 25 m of head at 8 L/s on its H-Q curve. The system requires 10 m of static head plus friction losses of 0.3·Q² meters (with Q in L/s). At what flow rate does this pump-system combination operate?

• A. 8 L/s — the pump always delivers its rated flow regardless of the system
• B. The operating point must be found graphically or algebraically; it is where the pump head equals the system head demand, not necessarily at the rated point
• C. 0 L/s — the static head of 10 m exceeds the pump's shutoff head
• D. The pump cannot operate because its rated head (25 m) is too high for the system

Answer: (B)

The operating point is not a property of the pump alone — it is the intersection of the pump H-Q curve with the system curve. The pump 'wants' to deliver various combinations of head and flow (defined by its curve); the system 'demands' more head as flow increases (static head + friction losses ∝ Q²). The actual flow rate is where supply equals demand: pump head = system head. This may or may not coincide with the pump's rated or best-efficiency point. Option A captures the most common misconception: students think the pump determines flow unilaterally, ignoring that the system curve constrains the operating point.

Question 2

A pump running at 1200 RPM consumes 80 kW. The operator uses a variable-frequency drive to reduce speed to 960 RPM. Approximately how much power does the pump now consume?

• A. 64 kW — power scales linearly with speed (960/1200 × 80)
• B. 51.2 kW — power scales as the square of the speed ratio ((960/1200)² × 80)
• C. 40.96 kW — power scales as the cube of the speed ratio ((960/1200)³ × 80)
• D. 80 kW — power is determined by the system demand, not the pump speed

Answer: (C)

The affinity laws state P ∝ N³. Speed ratio = 960/1200 = 0.8. New power = 80 × (0.8)³ = 80 × 0.512 = 40.96 kW. This cubic relationship is why variable-speed drives are so effective for energy savings: a modest speed reduction produces a dramatic power reduction. A 20% speed reduction cuts power consumption by nearly 49%. Option A (linear) and B (square) represent common partial-knowledge errors — students sometimes remember that Q ∝ N or H ∝ N² but forget the correct exponent for power.

Question 3

The operating point of a pump-piping system is determined solely by the pump's H-Q curve — it is the flow rate at which the pump reaches its maximum efficiency.

• T. True
• F. False

Answer: False

The operating point is determined by the intersection of two curves: the pump's H-Q curve (head supplied vs. flow) and the system curve (head demanded vs. flow). Neither curve alone determines the operating point. The system curve depends on static head (elevation difference) and dynamic losses (pipe friction, fittings), both of which are properties of the piping system, not the pump. Maximum efficiency is a property of the pump curve and may or may not coincide with the operating point — good system design tries to match them, but they are conceptually distinct.

Question 4

Reducing pump speed by 20% reduces power consumption by approximately 49%, because pump power scales as the cube of the speed ratio.

• T. True
• F. False

Answer: True

By the affinity law P ∝ N³: P₂/P₁ = (N₂/N₁)³ = (0.8)³ = 0.512. So the new power is 51.2% of the original — a reduction of 48.8%, approximately 49%. This cubic law is the quantitative basis for variable-speed drive energy savings. It also means the effect is asymmetric: a 20% speed increase multiplies power by (1.2)³ = 1.728, a 73% increase. Understanding this scaling is essential for energy-efficient pump selection and operation.

Question 5

Explain what cavitation is, why it damages pump impellers, and what conditions at the pump inlet cause it to occur.

Model answer: Cavitation occurs when local pressure in the pump inlet drops below the liquid's vapor pressure, causing the liquid to locally vaporize and form vapor bubbles. These bubbles are carried into higher-pressure regions of the impeller where they collapse violently, generating micro-shock waves that erode the impeller surface over time. The result is noise, vibration, reduced performance, and progressive mechanical damage. NPSH_available = (absolute inlet pressure head) − (vapor pressure head) quantifies the margin before cavitation. Cavitation occurs when NPSH_available falls below NPSH_required (the manufacturer's threshold). NPSH_available decreases with: high pump elevation above liquid source (increases suction lift), high suction pipe friction losses, high liquid temperature (increases vapor pressure), and high altitude operation (lower atmospheric pressure).

Cavitation is not simply 'boiling' — the collapse event is the damaging mechanism. Bubbles form in low-pressure zones and collapse in nanoseconds when they reach higher pressure, releasing energy that locally exceeds the yield strength of even hardened steel. Preventing it requires keeping NPSH_available > NPSH_required with a safety margin, which constrains pump placement, inlet pipe sizing, and operating conditions.